Is the impact of being hit by a bullet the same as the recoil felt by the shooter?

I’ve had a long running argument with both my girlfriend’s father as well as her brother regarding the force of a bullet striking a target. We usually say this target is a person, for the sake of easy discussion.

It is my general contention that the bullet strikes the target with the same amount of force that the shooter experiences as the firearm's recoil. We all recognize that Hollywood, in general, is full of **** (people being thrown across the room when shot is stupid). But they think that the target is hit with more energy than the shooter. I argue that the energy is the same, but that the bullet’s force is spread out over a very small contact area (the leading edge of the bullet), and therefore allows the bullet to penetrate the target. I argue that we experience a much smaller sensation of force because that force is absorbed by the weight of the gun, and spread out over our arm, through our shoulder, and into the rest of our body.

They make the (excellent sounding) counter argument that if what I was saying is true, then if we could somehow place a pistol right in front of someone’s chest, and allow the recoil to make it slam into that person’s chest, then it would feel very similar to the experience of wearing a bullet resistant vest and being hit with that same bullet in the same spot (more or less assuming that the force would be distributed in a similar amount of area). I agree with them that, from all accounts, it would probably be much worse to be hit by the bullet than be hit by the gun.

Does anyone have an opinion or insight into this? I looked up the physics on wikipedia, but I can’t interpret the meaning because it refers to momentum:

Assuming the gun and shooter are at rest, the force on the bullet is equal to that on the gun-shooter. This is due to Newton's third law of motion (For every action, there is an equal and opposite reaction). Consider a system where the gun and shooter have a combined mass M and the bullet has a mass m. When the gun is fired, the two systems move away from one another with new velocities V and v respectively. But the law of conservation of momentum states that the magnitudes of their momenta must be equal:

MV + mv = 0

Since force equals the rate of change in momentum and the initial momenta are zero, the force on the bullet must therefore be the same as the force on the gun/shooter.

Hollywood depictions of firearm victims being thrown through plate-glass windows are inaccurate. Were this to be the case, the shooter would also be thrown backwards with equal force. Gunshot victims frequently fall or collapse when shot; this is less a result of the momentum of the bullet pushing them over, but is primarily caused by physical damage or psychological effects, perhaps combined with being off-balance.

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As a quick aside concerning the math: The equation initially looked blatantly incorrect to me. But velocity is speed PLUS direction. Since the shooter and bullet are moving in two different directions, one of them has a negative velocity (one of them is moving backwards, depending on your perspective – probably the shooter in most people’s opinion). So if you give one of the systems a negative velocity, then multiply it by that system’s mass, then you get a negative value, which (according to Newton) is the inverse of the other system.

If my perspective holds out, then anyone being shot with just about any firearm should be able to stand up against the hit (at least until the internal damage makes you drop). By this rationale, you should even be able to keep your feet if hit in the chest by a 12 gauge (since the shooter is able keep standing while shooting). But this, again, seems unlikely. Of course, I have never seen anyone hit in the chest by a shotgun blast, so I can’t speak with any authority about this.

Please weigh in and let me know what you think (firsthand accounts of being, or seeing, someone shot are especially welcome).

(TL;DR Is the impact of being hit by a bullet the same as the recoil felt by the shooter?)

Yes, but because the surface area of transfer from the bullet to the target is so much smaller than the surface area of transfer from the gun to the shooter, it has a more severe (destructive) effect.
Some people will jump back when shot, as if the bullet was actually pushing them, but this is recognized to be a physiological response to pain, where the body is trying to throw itself clear (the same way that many people will involuntarily jump back if they touch something hot, or have a very bright light shone into their eyes). And, accounting for the slight loss of the bullet's energy while in flight, the energy transferred to the target would actually be slightly LESS than the energy felt by the shooter.

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Sort of. Remember "every action has an equal and opposite reaction". That law tells us that the backwards kick against the empty case is the same as the forwards push against the bullet. However, you have to modifiy that with two of the other laws of physics - "an object at rest tends to remain at rest" and the other one (which I can't quote) about the force required to move an object increases as the mass of the object increases.

That being said, yes the initial backwards recoil is exactly the same as the forward bullet push. But, much of that backwards energy is absorbed by the mass (weight) of the rifle before it hits your shoulder.

No, The force is not the same. The force of impact is simply the weight of the projectile x the velocity traveling. That is the impact. The purpose of the ballistic charts shows that the round reaches a maximum velocity at some point (usually very soon after leaving the muzzle), it is at that point that the maximum impact is available. It declines from there. The 'recoil' impact is a little more complex but I believe it is the velocity x weight of projectile into the surface area of the chamber less things like ejection mechanics etc.
I do believe in simplistic terms the impact force of a normal round can be no more than the force of recoil potential, and is usually less.

The simple answer is yes. The real world is very complex. The actual recoil energy is slightly lower than the energy of the bullet. However, the basic principle of equal and opposite is close enough.

__________________
Still happily answering to the call-sign Peetza.
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The problem, as you so eloquently put it, is choice.
-The Architect
-----
He is no fool who gives what he can not keep to gain what he can not lose.
-Jim Eliott, paraphrasing Philip Henry.

The short answer to your question is that the shooter and the recipient, assuming the shooter holds on to the gun and the recipient stops the bullet, is that they both experience the same total push.

However, the dissipated energy is very different for the two. A firearm expels a round by accelerating the round in one direction while using its own weight to "push off". You don't really feel the recoil until the bullet is long gone and the firearm continues moving towards you.

The firearm weighs an order of magnitude more than the bullet. You can expect them to have theoretically identical and realistically similar momentums since they separated as an enclosed system and with minimal external interference.

The difference comes along when you consider velocity. The force of the expanding powder pushed forward on the bullet and backwards on the gun over a finite period of time with a varying force (we won't get into differential equations this time around). This imparted identical momentum to the two units, but since the gun weighs an order of magnitude more than the bullet, the bullet was moving much faster. The faster, lighter bullet has far more kinetic energy than the heavy gun.

Momentum of the bullet = 1 * 500 = 500
The gun's momentum is also 500, by definition.
The gun's speed = 500 / 100, or 5 units of distance per second. This is a function of the ratio between the weight of the gun and the weight of the bullet and will always vary linearly.

Now, look at energy. The gun, moving at 5 units per second, has an energy of (.5)(500)(5^2), or 6250. The bullet has an energy of (.5)(1)(500^2), or 125000. In our simple example, that equates to 20 times the kinetic energy of the gun.

Without getting any crazier, you can look at an impact in terms of what is called the "impulse". The impulse can be displayed as a graphed line showing pressure over time. The slow, large gun makes a relatively long gentle impulse against your hands. The small, fast bullet makes a much sharper impulse against its target. The integral of both impulses (the area under the graph, representing total force to slow down the object) is the same in both cases although the shapes will be much different.

Torque is a measure of rotational force, so it's not quite the right term here. However, blocking a firearm such that it's unable to move would be the equivilent of making its weight infinite. There is a diminishing return in raising the weight of a firearm as most of the energy of the blast is transferred to the bullet anyway due to the large ratio.

In the above example, of 21 fractions of the total energy of the system, 20 went to the bullet and 1 went to the gun. If you doubled the weight of the gun, you could raise that to 20.5 and .5. If you made the gun infinitely heavy (I.E., secured it to a wall), you could transfer all 21 units to the round. The momentum of the round would increase by the same percentage.

I would say yes and no.
With the firearm, you have the buttstock/grip with all those cubic inches of surface area to disperse the momentum/force/whatever. With the bullet, you have a very small amount of surface area. Because of this you get penetration and subsequent terminal damage.

what MCCALL said +1 and add----surface area that is very hot.
the heat which transfers to the new 'medium' it finds itself in.
though it may be difficult to differentiate the being shot pain
from the being cooked pain.
me no wanna try.

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Make a fire for a man and you warm him for the night
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There really is no "yes and no". If you believe in the laws of physics then the answer is "Yes." I don't think the OP was referencing pain. The question is "energy transfer". The transfered energy, minus any losses to things like friction and noise, is identical.

The effect is different, the energy is identical.

__________________
Still happily answering to the call-sign Peetza.
---
The problem, as you so eloquently put it, is choice.
-The Architect
-----
He is no fool who gives what he can not keep to gain what he can not lose.
-Jim Eliott, paraphrasing Philip Henry.

The momentum of a 158 grain bullet in a 357 has been calculated to be the same as a 90 mph fast ball.

Batters have been killed by fastballs, which is why batters now wear helmets.

When I see a guy hit by a fastball, I see pain, I see crumpling, I see hobbling. I see a guy's own physical response, but I don't see him being blown to the backstop.

I have seen 200 pound men jump a country mile after being stung by a hornet. Seen them run real fast to get away. Was it the "energy" of the sting or the reaction of the person that caused the jump?

Obviously, it was the reaction of the person.

You can probably find videos of people wearing ballistic vests and being shot. From the images, looks like they are being punched hard and fast. I will bet some are bruised. I heard one News reporter broke a rib.

Momentum is a real killer. Something heavy moving slowly will cause real problems for soft shelled humans. Just drop a 50 pound sack of concrete mix on someone from 10 feet, 20 feet, or higher. I will bet a 10 foot drop will break a persons neck.

Peoples own responses to being shot vary wildly. I have read of guys thinking they were stung, then they looked down and found they were missing a foot.

Peetzakilla - the energy is most certainly -not- identical. See above. This is misinformation.

It most certainly is not. You are making assumptions about my statement that are not valid. Conservation of energy is an absolute, indisputable law of physics. Conservation of momentum can not be violated. The total energy of the system is a constant. The energy transfered by the bullet to the target is identical to the energy transfered to the gun and shooter, discounting loses than can all be essentially traced to friction and/or noise.

Your statement that "you are confusing momentum with energy" is misleading. Momentum IS energy. It's not mathematically the same as KINETIC energy but it most certainly IS energy.

In fact, assuming a perfectly inelastic collision, the kinetic energy is perfectly converted to other forms of energy and is equally conserved.

__________________
Still happily answering to the call-sign Peetza.
---
The problem, as you so eloquently put it, is choice.
-The Architect
-----
He is no fool who gives what he can not keep to gain what he can not lose.
-Jim Eliott, paraphrasing Philip Henry.

I'm not going to argue this any more. Momentum is not energy. I'm not reading into your words - allow me to quote your previous post to which I was replying:

Quote:

...

The effect is different, the energy is identical.

Momentum is a measure of inertia. While it can be indicative of energy, to say that it is energy is incorrect. Perhaps a physics class is in order. Terms describing different aspects of Newtonian physics are not interchangeable.

Peetza,
rsgraebert has it right on this one. Energy is conserved in the sense that all the potential energy of the powder is conserved (transfered to bullet, rifle and lost to the surrounding), but it is not distributed to the bullet and rifle equally.
Momentum is not energy...

ok, first of all, the answer is no. because there is air resistance, and the slim resistance in the barrel traveling to its target.
second, the force is "similar" but felt in a different way. You are feeling the force acting back at you from the recoil. But, you are holding a gun that is at least a pound in weight, and it is being distributed throughout your fingers, palms, arms, and so forth. You are absorbing the "shock" of the blast with more of a surface area than the target is. Way more
The bullet on the other hand on its traveling path, through air resistance, and other forms of resistance such as gravity(pulling it down at 9.8 feet/second) but it will impact in a MUCH smaller point of pressure per square inch than the recoil in your palms.
The best example is this classic question : Who is exerting more force per square inch, the fat lady in heels? Or the 2 ton elephant? The answer is the fat lady in heels. The back of the heel has to support all of her weight in that little .5 inch-1 inch heel boot. That is LOTS of pressure/square inch. The concept is the same for feeling recoil/bullet felt.
Caveat: Different people have different perceptions on what recoil really is. Lots to you, can be nothing to me.

__________________
1. The gun is always loaded.
2. Never let the muzzle cover anything you are not willing to destroy.
3. Keep your finger off the trigger unless you are ready to shoot.
4. Be be sure of your target and what is beyond it.

Momentum and kinetic energy are both measures of the quantity of motion, and a sideshow in the Newton-Leibnitz controversy over who invented calculus was an argument over whether mv (i.e., momentum) or mv2 (i.e., kinetic energy without the 1/2 in front) was the “true” measure of motion. The modern student can certainly be excused for wondering why we need both quantities, when their complementary nature was not evident to the greatest minds of the 1700's. The following table highlights their differences.

Example 4: A spinning top

A spinning top has zero total momentum, because for every moving point, there is another point on the opposite side that cancels its momentum. It does, however, have kinetic energy.

Example 5: Momentum and kinetic energy in firing a rifle

The rifle and bullet have zero momentum and zero kinetic energy to start with. When the trigger is pulled, the bullet gains some momentum in the forward direction, but this is canceled by the rifle's backward momentum, so the total momentum is still zero. The kinetic energies of the gun and bullet are both positive scalars, however, and do not cancel. The total kinetic energy is allowed to increase, because kinetic energy is being traded for other forms of energy. Initially there is chemical energy in the gunpowder. This chemical energy is converted into heat, sound, and kinetic energy. The gun's “backward”' kinetic energy does not refrigerate the shooter's shoulder!

__________________
Still happily answering to the call-sign Peetza.
---
The problem, as you so eloquently put it, is choice.
-The Architect
-----
He is no fool who gives what he can not keep to gain what he can not lose.
-Jim Eliott, paraphrasing Philip Henry.

The rifle and bullet have zero momentum and zero kinetic energy to start with. When the trigger is pulled, the bullet gains some momentum in the forward direction, but this is canceled by the rifle's backward momentum, so the total momentum is still zero.

Thanks for proving my point, because if two objects of different mass have the same momentum, the smaller object will have more kinetic energy...
That's just how it is....

Quote:

Momentum and kinetic energy are BOTH measures of energy.

And no it really isn't... The models for conservation of energy and conservation of momentum are different, but are often used in conjunction to represent a constitutive equation for motion, thus can both be used to represent the same thing. In other words, momentum can be a representation of energy, but it doesn't mean that it is energy...

About 10 years ago, there was a documentary titled something like, "How the Movies Get it Wrong". Surprisingly, much of the coverage was on firearms and explosives.

Part of the documentary covered the 'Knocked off your feet' myth. They did a few tests with bean bags, which had the same calculated energy levels as certain projectiles in question. The narrator discussed the mathematics and energy transfer involved, as some unfortunate volunteers got blasted with the bean bags. Some dropped to the floor; some didn't. It provided a nice segue into the next subject.

They talked to a Psychology PhD, that had performed a 5 year study of 'culture vs gun shot wounds'. The PhD had reviewed tens of thousands of hours of war films, street riots (with guns), and firefights; in which some of the subjects received gun shot wounds.

His first set of findings was fairly straight forward, mostly physiological, and easily understandable:
1. A hit (or hydrostatic shock) to the central nervous system usually causes a major disruption of bodily function. The subject almost always drops to the ground; and often loses consciousness, at least for a moment.
2. A hit to a major artery or the heart is immediately sensed by the body. (Even a hit to the femoral artery causes an instantaneous drop in blood pressure.) The body's reflexes often cause the subject to drop to the ground.

His second set of findings were the most interesting:
The key factor here, was whether or not the subject knew they had been shot.
In cultures that had a large Hollywood presence; even minor gunshot wounds resulted in the person hitting the ground.
Yet, in cultures that had zero, to minimal Hollywood presence; even major gunshot wounds did not cause them the drop.

The final bit of discussion with the PhD was about Americans, in particular. He had found that, even with broken bones, if the subject didn't know of their wound, they showed no signs of being affected. However, in many cases, the wound would become apparent in some way, and they would instantly hit the ground. The documentary showed a few examples of this. The wounded police officer or soldier was continuing with their duties, but would drop to the ground as soon as they were aware of the wound. It was like a light switch.

It all boils down to what the 'victim' believes they should do, when shot.
Emotional, physiological, and mechanical (broken bones) factors come into play.

---That being said....
Transfer of energy to a shooter vs the shootee is very different.
Many of the issues have already been pointed out, but I feel the most important ones are:
1. The weight of the firearm.
. A 6.5lb rifle is not going to accelerate as quickly as the projectile, and encounters resistance (your shoulder) almost immediately.
2. The surface area being impacted.
. Dispersing the recoil energy over a larger surface can 'rock' your body fairly easily, with certain cartridges and rifles. However, the opposing force is exerted by the projectile on a much smaller surface area. You get penetration, rather than a pushing force.
3. Firearm bores are often offset, above the stock or grip.
. Offset bores direct some of the recoil energy into a non-linear motion. By directing the motion upwards, rather than back, the shooter doesn't feel the full recoil force.

__________________
Don't even try it. It's even worse than the internet would lead you to believe.

Einstein and Newton would be shaking their heads at this discussion, LOL.

Since you guys have already hashed out all of the math, I'll just make a couple of comments to think about:

As far as bracing a rifle stock against a solid object, good stocks have been cracked, splintered and broken by the practice. The recipient of a speeding bullet is much like the braced rifle stock, as the target doesn't have time to move in response to the energy being dumped into it, thus much more damage is done on the down range end of the equation.

What would happen if you attached a 2 foot long hardened steel rod with a bullet shaped tip protruding out the back of a rifle stock, poked it against a dead pig & discharged the weapon? The penetration would be much less then on the dead pig downrange, who received the FMJ bullet @ 2,000+ fps BUT, they would both have absorbed roughly the same amount of energy.

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The Bible is my lawbook. I turn the other cheek when applicable, and spend the rest of my days resisting evil at every front, until I have breathed my last breath.

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